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viva [34]
3 years ago
12

Michael is the captain of his school’s soccer team. What skill does Michael exhibit when he decides which player will take the p

enalty kick?
A. conflict resolution
B. sportsmanship
C. leadership
D. teamwork
Physics
1 answer:
aleksley [76]3 years ago
5 0
A because if he was a leader he’d take the kick himself and they’re are not all working to kick the ball so it’s not teamwork and it’s not sportsmanship because he isn’t stopping a fight or fixing something between two teams
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Two atoms that are isotopes will have a different number of what​
BigorU [14]

Answer:

different number of mass numbers.

Explanation:

isotopes are atoms of the same element having the same atomic number but different mass numbers due to different number of neutrons.

7 0
3 years ago
A sample of water with a mass of 1 gram freezes at 32°F. At what temperature will a 100-gram sample of water freeze?
valkas [14]

Answer:

3200°F

Explanation:

just add two zero's to the end

8 0
3 years ago
An open system starts with 52 J of mechanical energy. The energy changes
Vikki [24]

Answer:

47 J

hope it helps u

thanks for easy ask

3 0
2 years ago
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl
Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0
2 years ago
Which sentence states Newton’s third law?
hammer [34]
<span>If two objects collide, each object exerts a force equal to and in the opposite direction of the other.</span>
3 0
2 years ago
Read 2 more answers
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